Ask question

Ask question

All categories

3 years ago

14

In 1999, there were 9860 Great Danes registered with the American Kennel Club. The number of registered Labrador Retrievers was

6997 more than fifteen times the number of registered Great Danes. How many registered Labrador Retrievers were there?
show your work please

2 answers:

katovenus [111]3 years ago

8 0

Answer:

154,897 REGISTERED LABRADOR RETRIEVERS WERE THERE

Step-by-step explanation:

Given the statement: IN 1999, THERE WERE 9860 GREAT DANES REGISTERED WITH THE AMERICAN KENNEL CLUB.

Also, THE NUMBER OF REGISTERED LABRADOR RETRIEVERS WAS 6997 MORE THAN FIFTEEN TIMES THE NUMBER OF REGISTERED GREAT DANES

Let x represents the number of Great Danes registered and y represents the number of Labrador retrievers registered.

The statement:

"Fifteen times the number of registered great Danes " means 15x

and

"6997 more than fifteen times the number of registered great Danes" means 15x + 6997

then as per the given statement:

x = 9860 .

To solve for y;

y = 15x+ 6997

Substitute the value of x we get;

Simplify:

y = 154,897

Therefore, the registered Labrador Retrievers were there are; 154,897

douwdek0 and 16 more users found this answer helpful

7nadin3 [17]3 years ago

5 0

Eeeeeeeeeeeeeeeeerreeeeeeeeeerrrererrrrrrrre

You might be interested in

1.In what ways did Egypt prosper during the reign of Hatshepsut? do your thing brainly!

Bond [772]

Answer:

She served as a queen, for Thutmose ll.

She served as queen alongside her husband, Thutmose II, but after his death, she claimed the role of pharaoh!. She reigned peace, she built temples and monuments, resulting in the flourishing of Egypt!

srry if this didn't help but I hope it did!

6 0

3 years ago

From a radar station, the angle of elevation of an approaching airplane is 32.5 degree. The horizontal distance between the plan

denis23 [38]

Answer: 42.21 km

Step-by-step explanation:

We can solve this using trigonometry, since we have the following data:

$\theta=32.5\°$ is the the angle of elevation

$d=35.6 km$ is the horizontal distance between the plane and the radar station

$x$ is the hypotenuse of the right triangle formed between the radar station and the airplane

Now, the trigonometric function that will be used is <u>cosine</u>:

$cos\theta=\frac{d}{x}$ because $d$ is the adjacent side of the right triangle

$cos(32.5\°)=\frac{35.6 km}{x}$

Finding $x$ :

$x=\frac{35.6 km}{cos(32.5\°)}$

$x=42.21 km$

4 0

3 years ago

Omgggg help me for brainly what’s 4 x 5 divided by 2

yanalaym [24]

Answer:

10

Step-by-step explanation:

$\frac{4 x 5}{2}$

4 x 5 = 20

$\frac{20}{2}$ = 10

3 0

3 years ago

Read 2 more answers

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

3 years ago

Question 3 Unsaved

sashaice [31]

87.4 people per square mile

3 0

3 years ago

Read 2 more answers